Problem: Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $k \neq 0$. $p = \dfrac{4k^2 - 32k - 36}{2k^2 - 2k - 4} \times \dfrac{4k - 8}{2k + 6} $
First factor out any common factors. $p = \dfrac{4(k^2 - 8k - 9)}{2(k^2 - k - 2)} \times \dfrac{4(k - 2)}{2(k + 3)} $ Then factor the quadratic expressions. $p = \dfrac {4(k + 1)(k - 9)} {2(k + 1)(k - 2)} \times \dfrac {4(k - 2)} {2(k + 3)} $ Then multiply the two numerators and multiply the two denominators. $p = \dfrac { 4(k + 1)(k - 9) \times 4(k - 2)} { 2(k + 1)(k - 2) \times 2(k + 3)} $ $p = \dfrac {16(k + 1)(k - 9)(k - 2)} {4(k + 1)(k - 2)(k + 3)} $ Notice that $(k + 1)$ and $(k - 2)$ appear in both the numerator and denominator so we can cancel them. $p = \dfrac {16\cancel{(k + 1)}(k - 9)(k - 2)} {4\cancel{(k + 1)}(k - 2)(k + 3)} $ We are dividing by $k + 1$ , so $k + 1 \neq 0$ Therefore, $k \neq -1$ $p = \dfrac {16\cancel{(k + 1)}(k - 9)\cancel{(k - 2)}} {4\cancel{(k + 1)}\cancel{(k - 2)}(k + 3)} $ We are dividing by $k - 2$ , so $k - 2 \neq 0$ Therefore, $k \neq 2$ $p = \dfrac {16(k - 9)} {4(k + 3)} $ $ p = \dfrac{4(k - 9)}{k + 3}; k \neq -1; k \neq 2 $